Maxima and Minima of functions of single variable

MAXIMA AND MINIMA OF FUNCTIONS OF SINGLE VARIABLE

 

The important application of differential calculus are optimization problems, in which we are required to find the optimal (best) way of doing the problems.

 

 

1. APPLIED MAXIMA AND MINIMA

 

To find the maxima and minima of a single‒variable function y = f(x),

Find the first derivative: Calculate the first derivative of the function, dy/dx.

Find critical points: Set dy/dx = 0 and solve for x to find the critical points.

Find the second derivative: Calculate the second derivative of the function, denoted as d²y/dx².

 

Evaluate the second derivative at critical points:

Maximum: If d²y/dx² < 0 at a critical point, the function has a maximum at that point.

Minimum: If d²y/dx² > 0 at a critical point, the function has a minimum at that point.

Inconclusive: If d²y/dx² = 0 at a critical point, the test is inconclusive, and you must use the first derivative test.

 

Procedure for Solving Applied Maximum and Minimum Problems

Step 1. Draw an appropriate figure and label the quantities relevant to the problem.

Step 2. Find a formula for the quantity to be maximized or minimized.

Step 3. Using the conditions stated in the problem to eliminate variables, express the quantity to be maximized or minimized as a function of one variable.

Step 4. Find the interval of possible values for this variable from the physical

restrictions in the problem.

Step 5. If applicable, use the techniques to obtain the maximum or minimum.

 

 

Example 126. A garden is to be laid out in a rectangular area and protected by a chicken wire fence. What is the largest possible area of the garden if only 100 running feet of chicken wire is available for the fence?

Solution:

Let x = length of the rectangle (ft)

y = width of the rectangle (ft)

A = area of the rectangle (ft2)

Then,

A = xy... (1)

Since the perimeter of the rectangle is 100 ft, the variables x and y are related by the equation

2x + 2y = 100 or y = 50‒x   ... (2)

Substituting (2) in (1) yields

A = x(50 − x) = 50x ‒ x²   ... (3)

Because x represents a length, it cannot be negative, from (3)x cannot exceed 50,

0≤ x ≤ 50       . . . (4)

Thus, we have reduced the problem to that of finding the value (or values) of x in [0,50], for which A is maximum. Since A is a polynomial in x, it is continuous on [0,50], and so the maximum must occur at an endpoint of this interval or at a critical point.

From (3) we obtain

dA/dx =50‒2x

For maximum

dA/dx = 0

50‒2x=0

x = 25

Thus, the maximum occurs at one of the values

 x = 0, x = 25, x = 50

Substituting these values in (3) yields, which tells us that the maximum area of 625 ft² occurs at x=25, which is consistent with the graph of (3). From (2) the corresponding value of y is 25, so the rectangle of perimeter 100 ft with greatest area is a square with sides of length 25 ft.

 

Example 127. An open box is to be made from a 16‒inch by 30‒inch piece of card‒board by cutting out squares of equal size from the four corners and bending up the sides. What size should the squares be to obtain a box with the

largest volume?

Solution: The cardboard piece with squares removed from its corners.

Let x = length (in inches) of the sides of the squares to be cut out

V = volume (in cubic inches) of the resulting box

Because we are removing a square of side x from each corner, the resulting box will have dimensions 16‒2x, 30‒2x, x

Since the volume of a box is the product of its dimensions, we have

V = (16 ‒ 2x)(30 ‒ 2x)x = 480x ‒ 92x² + 4x³        ……(1)

Note that our volume expression is already in terms of the single variable x.

The variable x in (1) is subject to certain restrictions. Because x represents a length, it cannot be negative, and because the width of the cardboard is 16 inches, we cannot cut out squares whose sides are more than 8 inches long. Thus, the variable x in (1) must satisfy

0≤x≤8

and hence we have reduced our problem to finding the value (or values) of x in the interval [0, 8] for which (1) is a maximum.

From (1) we obtain

dV/dx = 480‒184x + 12x² = 4(3x² ‒ 46x + 120)

=4(x‒12)(3x‒10)

For maximum

dV/dx = 0

4(x‒12)(3x‒10) = 0

x=10/3 or x = 12

Since x = 12 falls outside the interval [0, 8], the maximum value of V occurs either at the critical point x =10/3 or at the endpoints x=0, x = 8. Substituting these values into (1) yields the Table given below, which tells us that the greatest possible volume V = 19,600 / 27 in³ ≈ 726 in³ occurs when we cut out squares whose sides have length 10/3 inches. This is consistent with the graph of (1).

 

Example 128. Find the radius and height of the right circular cylinder of largest volume that can be inscribed in a right circular cone with radius 6 inches and height 10 inches.

Solution: Let

r = radius (in inches) of the cylinder

h = height (in inches) of the cylinder

V = volume (in cubic inches) of the cylinder

The formula for the volume of the inscribed cylinder is

 V = πr²h       ... (1)

To eliminate one of the variables in (1) we need a relationship between r and h. The triangles ADE and ABC are similar.

we have

(10‒h) / r = 10/6

 ⇒  h = 10‒(5/3)r      ... (2)

Substituting (2) into (1) we obtain

V = πr² ( 10‒(10/3)r ) = 10πr² ‒ (5/3)πr³  ... (3)

which expresses V in terms of r alone. Because r represents a radius, it cannot be negative, and because the radius of the inscribed cylinder cannot exceed the radius of the cone, the variable r must satisfy

0 <r≤6

Thus, we have reduced the problem to that of finding the value (or values) of r in [0,6] for which (3) is a maximum. Since V is a continuous function of r on [0,6], the methods developed in the preceding section apply.

From (3) we obtain

 dV/dr = 20πr ‒ 5πг² = 5πr (4‒r)

For maximum

dV/dr = 0

5πr(4‒r) = 0

so r = 0 and r = 4 are critical points. Since these lie in the interval [0,6], the maximum must occur at one of the values

 r = 0, r = 4, r = 6

Substituting these values into (3) yields, which tells us that the maximum volume V = 160/3 π ≈  168 in³ occurs when the inscribed cylinder has radius 4 in. When r = 4 it follows from (2) that h = 10/3. Thus, the inscribed cylinder of largest volume has radius r = 4 inches and height h = 10/3 inches

 

Example 129. A window has the form of a rectangle surmounted by a semicircle. If the perimeter is 40 ft., find its dimensions so that greatest amount of light may be admitted.

Solution:

The greatest amount of light may be admitted means that the area of the window may be maximum.

Let ft. be the radius of the semi‒circle so that one side of the rectangle is 2x ft.

Let the other side of the rectangle be y ft.

Then the perimeter of the whole figure

 = perimeter of the semicircle + sum of 3 sides of the rectangle

 = πx + 2у + 2x     ... (1)

The area of the whole figure

A = area of the semicircle + area of the rectangle

= ½ πx² + length x breath

A = ½ πx² + 2xу    ... (2)

Given perimeter of the window is 40 ft.

 :. πx + 2y + 2x = 40       [From (1)]

 ⇒ 2у= 40‒πx ‒ 2x

 y = 20 ‒ πx/2 ‒ x  ... (3)

Substitute the value of y in (2)

A = ½πx² + 2x (20 ‒ πx/2 ‒ x)

 = ½πx² + 40x ‒ πx² ‒ 2x² = 40x ½πx² ‒ 2x²

dA/dx = 40‒πx ‒ 4x

d²A/ dx²  = ‒π‒4 < 0

For maximum

dA/dx = 0

40‒πx‒4x=0

 ‒πx‒4x= ‒40

 x(π+4)= 40

 x = 40 / π+4

 x = 5.6 ft.

Substitute the value of x in (3) to find y,

 y=20 ‒ [π(5.6)/2] ‒ 5.6 = 5.608 ft.

The dimensions are 2x and y

11.2 ft and 5.608 ft

 

 

2. ABSOLUTE MAXIMUM AND MINIMUM

 

Definition: Let c be the number in the domain D of a function f.

Then f(c) is the,

a) Absolute maximum value of f on D if f(c) ≥ f(x) for all x ∈ D

b) Absolute minimum value of f on D if f(c) ≤ f(x) for all x ∈ D

An absolute maximum or minimum values is called a global maximum or minimum. The maximum and minimum values of ƒ are called extreme values of f.

The above diagram shows the graph of a function f absolute maximum at d and absolute minimum at a. Note that, (d, f(d)) is the highest point on the graph and (a, (ƒ(a)) is the lowest point. If we consider only value of x near b then f(b) is the largest of those values of f(x) and is called a local maximum value of f. f(c) is called a local minimum value of ƒ because ƒ (c) ≤ f(x) for x near c. The function ƒ also has a local minimum at e.

 

Definition: Critical point

A number c is said to be critical point of f (x) on D, If f'(c) = 0

Procedure for finding the absolute minima and the absolute maxima

To find the absolute maxima or absolute minima:

• Let f (x) be a function defined on the interval [a, b].

• Take the critical points which lies in [a, b]. Let them be x₁, x₂,...x_n.

• Find f(a), f(b), f(x₁), ƒ (x₂), ... ƒ (x_n)

• From the above, the value minimum value is absolute minima and the maximum value is absolute maxima

 

Example 130. Find the critical points for (x) = 5x³‒6x+1.

Solution:

The given function

f(x) = 5x³ ‒ 6x + 1

f '(x) = 15x² ‒ 6

To find the critical point:

f '(x) = 0

15x²‒6=0

3(5x²‒2) = 0

⇒ 5x²‒2=0

⇒ x = ±√(2/5)

 

Example 131. Find the absolute maximum and minimum for f(x) = 3x⁴ ‒ 4x³‒ 12x² + 1 on [−2,3]

Solution: The given function

 f(x)=3x⁴‒4x³‒12x²+1

is a polynomial, which is continuous on [‒2,3].

 f '(x) = 12x³‒12x² ‒ 24x

= 12x(x²‒x‒2)

= 12x(x‒2)(x+1)

To find the critical point:

f '(x) = 0

12x(x‒2)(x+1)= 0

⇒x=0,2,‒1

Each of this critical points lies in the interval [‒2,3].

The value of the function at these critical points are

f(0) = 3(0) ‒ 4(0) – 12(0) + 1 = 1

f(2) = 3(2)⁴ ‒ 4(2)³ ‒ 12(2)² + 1

= 48‒32‒48+1= ‒31

f(‒1)=3(‒1)⁴ — 4(‒1)³ ‒ 12(‒1)² + 1

= 3+4 ‒ 12+ 1 = −4

The value of the function at the end points of the interval [‒2,3] are

f(‒2) = 3x⁴ ‒ 4x³ ‒ 12x² + 1

= 3(‒2)⁴ ‒ 4(‒2)³ ‒ 12(‒2)² + 1

= 48+32‒48 + 1 = 33

f(3) = 3x⁴ ‒ 4x³ – 12x² + 1

= 3(3)⁴ ‒ 4(3)³ ‒ 12(3)² + 1

= 3(81) ‒ 4(27) ‒ 12(9) + 1

= 243‒108‒108+1 = 28

Comparing these five values, we see that

The absolute minimum is f(2) = ‒31

The absolute maximum of ƒ (3) = 33

 

Example 132. Find the absolute maximum and minimum for f(x) = x³‒3x²+1 on 1 on [‒1/2,4]

Solution: The given function

f(x) = x³‒3x² + 1

is a polynomial, which is continuous on [‒1/2,4]

f '(x) = 3x²‒6x

= 3x(x‒2)

To find the critical point:

f '(x) = 0

3x(x‒2)=0

⇒x=0, x=2 are critical points

Each of this critical points lies in the interval [‒1/2,4].

The value of the function at these critical points are

f(0) = 1

f(2) = (2)³‒3(2)² + 1

=8‒12+1

=9‒12

= ‒3

The value of the function at the end points of the interval [‒1/2, 4] are

f(‒1/2)=(‒1/2)³‒3(‒1/2)²+1

= ‒ 1/8 ‒ 3/4 ‒ 1

= [‒1‒6‒8] / 8 = ‒ 15/8

ƒ(4) = (4)³ − 3(4)² + 1

= 64 ‒ 3(16) + 1 = 65‒48

= 17

Comparing these four values, we see that

The absolute minimum of f(x) is f(2) = −3

The absolute maximum of f(x) is ƒ(4) = 17

 

Example 133. Find the absolute maximum and minimum for f(x) = (x² ‒ 1)³ on [‒1,2].

Solution: The given function

f(x) = (x² ‒ 1)³

is a polynomial, which is continuous on [‒1,2].

 f '(x)=3(x² ‒ 1)² 2x

= 6x(x²‒1)²

= 6x[(x + 1)(x − 1) ]²

To find the critical point:

 f '(x) = 0

6x[(x + 1)(x‒1)]² = 0

 ⇒ x=0,‒1,1 are critical points

Each of this critical points lies in the interval [‒1,2].

The value of the function at these critical points are

ƒ(0) = ((0)² ‒ 1)³

= (0‒1)³ = (‒1)³ = ‒1

f(‒1)=((‒1)² ‒ 1)³ = (1‒1)³ = 0

f(1) = ((1)² ‒ 1)³ = (1‒1)³ = 0

The value of the function at the end points of the interval [‒1, 2] are

ƒ(‒1) = ((‒1)² ‒ 1)³ = (1‒1)³ = 0

f(2) = ((2)² ‒ 1)³ = (4‒1)³ = 27

Comparing these four values, we see that

The absolute minimum of f(x) is f(0) = ‒1

The absolute maximum of f (x) is f(2) = 27

 

Example 134. Find the absolute maximum and minimum for f(x) = x + 1/x on [0.2,4]

Solution: The given function

f(x) = x + 1/x

is continuous on [0.2, 4].

f(x) = x+x^‒1

f '(x) = 1‒x^‒2

= 1 ‒ 1/x²

To find the critical point:

 f '(x) = 0

1 ‒ 1/x² = 0

1 = 1/x²

x² = 1

x = ±1

The critical point x = 1 lies in the interval [0.2, 4].

The value of the function at the critical point is

f(1) = 1+ 1/1 =2

The value of the function at the end points of the interval [0.2, 4] are

 f(0.2) = 0.2 + 1/0.2 = 0.2 + 5 = 5.2

 f(4) = 4 + ¼ = 4.25

Comparing these three values

The absolute minimum of f(x) is f(1) = 2

The absolute maximum of f(x) is f(0.2) = 5.2

 

Example 135. Find the absolute maximum and minimum for f(x) = x ‒ 2 sin x on [0,2π]

Solution: The given function

f(x) = x‒2 sin x

is continuous on [0,2π]

 f '(x) = 1 ‒ 2cos x

To find the critical point:

 f '(x) = 0

1‒2cosx = 0

2 cos x = 1

⇒ cosx = 1/2

x = cos^‒1(1/2)

x = π/3, 2π ‒ π/3

= π/3, 5π/3

Each of this critical points lies in the interval [0,2π].

The value of the function at these critical points are

 f(π/3) = π/3 ‒ 2sin(π/3)

= π/3 ‒ 2 × √3/2

= π/3 ‒  √3

= 3.14/3  ‒  √3

= ‒ 0.684

f(5π/3) = 5π/3 ‒ 2sin(5π/3)

= 5π/3 ‒ 2 × (‒√3/2)

= 5.23 ‒ 2(‒0.866)

= 6.968

The value of the function at the end points of the interval [0, 2π] are

f(0) = 0 ‒ 2(sin 0) = 0

f(2π) = 2π ‒ 2sin(2π) = 2π‒0 = 2 × 3.14

= 6.28

Comparing these four values

The absolute minimum of ƒ(x) is ƒ(π/3) = −0.68

The absolute maximum of ƒ(x) is ƒ(5π/3) = 6.968

 

Example 136. Find the absolute maximum and minimum for f(x)= x ‒ log x on [1/2,2].

Solution: The given function

f(x) = x ‒ logx

is continuous on [1/2,2]

ƒ'(x) = 1 – 1/x

To find the critical point:

 f '(x) = 0

1 ‒ 1/x = 0

1/x = 1

x = 1

The critical point x = 1 lies in the interval [1/2,2].

The value of the function at the critical point is

 f(1) = 1 ‒ log(1) = 1

The value of the function at the end points of the interval [1/2,2] are

f(1/2) = 1/2 ‒ log(1/2)

= 0.5+ (0.693) = 1.193

ƒ(2) = 2 – log(2) = 2 ‒ (0.693)

= 1.306

Comparing the three values

The absolute minimum of f(x) is f(1) = 1

The absolute maximum of f(x) is f(2) = 1.306

 

Example 137. Find the absolute maximum and minimum values of the function f(x) = log(x² + x + 1) in [−1, 1].

Solution: The given function

f(x) = log (x² + x + 1)

is continuous on [‒1,1]

f '(x) = 2x+1 / x²+x+1

To find the critical point:

 f '(x) = 0

2x+1 / x²+x+1  = 0

⇒2x+1=0

⇒ x = ‒1/2

The critical point x = ‒1/2 lies in the interval [−1, 1].

The value of the function at the critical point is

 f(‒1/2) = log_e (1/4 ‒ 1/2 + 1) = log_e(3/4)

= ‒ 0.288

The value of the function at the end points of the interval [‒1, 1] are

f(‒1)= log_e(1‒1+1) = log_e(1) = 0

f(1) = log_e(1 + 1 + 1) = log_e(3) = 1.099

Comparing the three values

The absolute minimum of f(x) is f(‒1/2) = ‒0.288

The absolute maximum of f(x) is f(1) = 1.099 

 

EXERCISE

 

34. Find the critical points for the following

a. f(x) = 2x³‒3x² ‒ 36            Ans: 3,‒2

b. f(x) = 2x ‒ 3x³            Ans: √2/3

 

35. Find the absolute maximum and minimum for f(x) = 2x³ ‒ 3x² ‒ 12x + 1 on

[‒2,3].     

Ans: The absolute minimum of f(x) is ƒ(−1) = −19

The absolute maximum of f(x) is f(2) = 8

 

36. Find the absolute maximum and minimum for f(x) = 3x⁴ ‒ 16x³ + 18x² on [‒1,4].

Ans: The absolute minimum of f(x) is ƒ(3) = −27

The absolute maximum of f(x) is f(−1) = 37

 

37. Find the absolute maxima and absolute minima for the following

a. f(x) = x³ − 3x² + 1 in [‒1/2,4]

Ans: Minimum value f(2) = ‒3, Maximum value f(4) = 17

b. f(x) = x³‒ 12x + 1 in [‒3,5]

Ans: Minimum value f(2) = ‒15, Maximum value ƒ(5) = 66

c. f(x) = x^a(1‒x)^b in [0,1]

Ans: Minimum value ƒ (0) = f(1) = 0, Maximum value f( a / a+b ) = a^ab^b / (a+b)^a+b

 

 

3. LOCAL MAXIMA AND LOCAL MINIMA

 

Consider the following graph

In the interval [a, b], f(x) increasing function. In the interval [b, c], f(x) is decreasing function. In the interval [c, d], f(x) is increasing function.

The graph shown in Fig. rises from A to B, falls from B to C, and rises again from C to D. The function f is said to be increasing on the interval [a, b], decreasing on [b, c], and increasing again on [c, d]. Notice that if x₁ and x₂ are any two numbers between a and b with x₁ < x₂, then f(x₁) < ƒ(x₂).

From the above concept, we can define the increasing and decreasing functions as given below:

 

Increasing function

A function f  is called increasing function on an interval I, if f(x₁) < f(x₂) where as x₁ < x₂

Decreasing function

A function ƒ is called decreasing function on an interval I, if f(x₁) > f(x₂) where as x₁ < x₂

Example:

Consider the following example f(x) = x² for x ∈ R.

In the interval (‒∞, 0], the function is decreasing and in the interval [0, ∞), the function is increasing.

Let c be that point in a domain D on a function f, then f(c) is

• local maximum value of f, if f(c) ≥ f(x) when x is near c.

• local minimum value of f, if f(c) ≤ f(x) when x is near c.

Consider the following figure.

The absolute minimum = f(a)

The absolute maximum= f(d)

f(x) is increasing in [a, b]

f(x) is decreasing in [b, c]

f(x) is increasing in [c, d]

f(x) is decreasing in [d, e]

In [a, c], the local minimum is f (a) and the local maximum is f (b)

In [d, e], the local maximum is f(d) and the local minimum is f(e)

The following figure shows local and absolute maximums occur at a single point and similarly shows the local and absolute minimums occur at a single point.